In John M. Lee's Introduction to smooth manifolds on p. 217 he proves the equality
$$d(\omega \wedge \mu) = d\omega\wedge \mu + (-1)^k\omega \wedge d\mu$$
I just don't understand a step in his proof and I can't get my head around it:
The crucial part is here:
$$(g\,df+f\,dg)\wedge dx^I\wedge dx^J = (df\wedge dx^I) \wedge (g\,dx^J) + (-1)^k(f\,dx^I)\wedge (dg\wedge dx^J)$$
I simply don't understand how we obtain the right hand side. What I did was:
$$(g\,df+f\,dg)\wedge dx^I\wedge dx^J = (g\,df)dx^I\wedge dx^J + (f\,dg)dx^I\wedge dx^J$$
How exactly do I get from my right hand side to the desired one? I know about the anticommutativity of the wedge product, but I just don't get why
$(g\,df)dx^I\wedge dx^J$ becomes (or is equal to) $(df\wedge dx^I) \wedge (g\,dx^J)$
whereas
$(f\,dg)dx^I\wedge dx^J$ all of a sudden takes care of the anticommutativity and turns into $(-1)^k(f\,dx^I)\wedge (dg\wedge dx^J)$
Any help is appreciated.
As @Ted Shifrin mentioned, switching a 1-form and a $k$-form introduces $(-1)^k$. In general, if $\omega$ is a $k$-form and $\mu$ is a $l$-form, then $$\omega\wedge\mu=(-1)^{kl}\mu\wedge\omega.$$
We can show it for $\omega=dx_1\wedge\cdots\wedge dx_k$ and $\mu=dx_{k+1}\wedge\cdots\wedge dx_{k+l}$ and extend linearly.
To switch from $\omega\wedge\mu$ to $\mu\wedge\omega$, we have to pass $dx_k$ through all $l$ factors of $\mu$, so we get $$\omega\wedge\mu=(-1)^l dx_1\wedge\cdots\wedge dx_{k-1}\wedge \mu\wedge dx_k$$
We repeat this for $k$ many factors, so $\omega\wedge\mu=(-1)^{kl}\mu\wedge\omega.$